# $12/15/2020$ - Daily Challenge Solution

$\underline{\text{Equations}}$

\begin{aligned} a + e &= 4 \\ b + i &= 5 \\ f + j &= 6 \\ g + k &= 7 \\ c + l &= 8 \\ d + h &= 9 \\ a + b + c + d &= 12 \\ e + f + g + h &= 13 \\ i + j + k + l &= 14 \end{aligned}

$\underline{\text{Brute-Force Case Work}}$

$a$ can be either one of $1, 2$ and $3$.

$d$ can be either one of $3, 4, 5$ and $6$.

Thus, their possible sums are $4, 5, 6, 7, 8$ and $9$.

$b$ can be either one of $1, 2, 3$ and $4$.

$c$ can be either one of $2, 3, 5$ and $6$.

Thus, their possible sums are $3, 4, 5, 6, 7, 8, 9$ and $10$.

(Let $|$ be the same as 'or'.)

So, $a + b + c + d$ is the same as $[4 | 5 | 6 | 7 | 8 | 9] + [3 | 4 | 5 | 6 | 7 | 8 | 9 | 10]$

The only pairs here that make $12$ are $[3 + 9], [4 + 8], [5 + 7] \text{ and } [6 + 6]$

Meh, this is getting nowhere...

Note by Percy Jackson
5 months ago

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# INCOMPLETE (and really frustrating...)

- 4 months, 4 weeks ago

Haha bro, I hate problems like those. No point in solving them lol

- 4 months, 3 weeks ago

Yes...I abandoned this when I started getting hopeless lol

- 4 months, 3 weeks ago

It turns out this problem is impossible. If the numbers from 1-10 are only allowed to be used ONCE, then we cannot solve column 1 or 2.

Comment if you want coding proof

- 4 months ago